带修改的一个区间分成几个极大连续区间
CDQ分治
1 #include <bits/stdc++.h> 2 3 using namespace std; 4 5 const int N = 2e5 + 10; 6 7 struct query{ 8 int x, y, val, op; 9 }q[N * 5], tmp[N * 5]; 10 int tot; 11 12 int n, m; 13 int a[N]; 14 int ans[N], idx; 15 int c[N]; 16 17 int lowbit(int x) 18 { 19 return x &(-x); 20 } 21 22 void add(int x, int y) 23 { 24 for (; x < N; x += lowbit(x)) 25 c[x] += y; 26 } 27 28 int ask(int x) 29 { 30 int res = 0; 31 for (; x; x -= lowbit(x)) 32 res += c[x]; 33 return res; 34 } 35 void clear(int x){ 36 for (; x < N; x += lowbit(x)) 37 c[x] = 0; 38 } 39 40 void cdq(int l, int r) 41 { 42 if(l == r) return ; 43 int mid = l + r >> 1; 44 cdq(l, mid), cdq(mid + 1, r); 45 int i = l, j = mid + 1, k = l; 46 while(i <= mid && j <= r) 47 { 48 if(q[i].x <= q[j].x) 49 { 50 if(!q[i].op) 51 { 52 add(q[i].y, q[i].val); 53 } 54 tmp[k ++] = q[i ++]; 55 } 56 else 57 { 58 if(q[j].op) ans[q[j].op] += ask(q[j].y) * q[j].val; 59 tmp[k ++] = q[j ++]; 60 } 61 } 62 while(i <= mid) 63 { 64 if(!q[i].op) add(q[i].y, q[i].val); 65 tmp[k ++] = q[i ++]; 66 } 67 while(j <= r) 68 { 69 if(q[j].op) ans[q[j].op] += ask(q[j].y) * q[j].val; 70 tmp[k ++] = q[j ++]; 71 } 72 for (int i = l; i <= r; i ++) 73 { 74 if(!q[i].op) clear(q[i].y); 75 q[i] = tmp[i]; 76 } 77 } 78 79 int main() 80 { 81 scanf("%d%d", &n, &m); 82 for (int i = 1; i <= n; i ++) 83 scanf("%d", &a[i]); 84 for (int i = 1; i <= n; i ++) 85 if(a[i] != a[i - 1]) 86 q[++ tot] = {i, a[i], 1, 0}; 87 int op,pos, v, x, y, l, r; 88 for (int i = 1; i <= m; i ++) 89 { 90 scanf("%d", &op); 91 if(op == 1) 92 { 93 scanf("%d%d", &pos, &v); 94 if(a[pos] == v) continue; 95 if(a[pos] != a[pos - 1]) 96 q[++ tot] = {pos, a[pos], -1, 0}; 97 if(a[pos] != a[pos + 1] && pos + 1 <= n) 98 q[++ tot] = {pos + 1, a[pos + 1], -1, 0}; 99 a[pos] = v; 100 if(a[pos] != a[pos - 1]) 101 q[++ tot] = {pos, a[pos], 1, 0}; 102 if(a[pos] != a[pos + 1] && pos + 1 <= n) 103 q[++ tot] = {pos + 1, a[pos + 1], 1, 0}; 104 } 105 else 106 { 107 scanf("%d%d%d%d", &l, &r, &x, &y); 108 ++ idx; 109 if(a[l] >= x && a[l] <= y) ++ ans[idx]; 110 l ++; 111 if(l > r) continue; 112 q[++ tot] = {r, y, 1, idx}; 113 q[++ tot] = {l - 1, y, -1, idx}; 114 if(x > 1) q[++ tot] = {r, x - 1, -1, idx}; 115 if(x > 1) q[++ tot] = {l - 1, x - 1, 1, idx}; 116 } 117 } 118 cdq(1, tot); 119 for (int i = 1; i <= idx; i ++) 120 printf("%d ", ans[i]); 121 }